Unit 8 Fluids: Full Practice Set

How to use this set

This set is written in the style of the AP Physics 1 exam. Every question is new and made for this site.

Section I: Multiple choice (18 questions)

Pick one answer for each question.

1. A small block has a mass of 0.30 kg and a volume of 4.0 × 10⁻⁴ m³. What is the density of the block?

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(B). ρ = m / V = 0.30 kg ÷ (4.0 × 10⁻⁴ m³) = 750 kg/m³.

(A) slips a power of ten. (C) divides V by m instead of m by V (1 ÷ 750 = 1.3 × 10⁻³, then a power-of-ten slip); check that kg ÷ m³ gives a sensible number. (D) multiplies m × V instead of dividing.

2. Two solid cubes are cut from the same block of plastic. Cube B has sides twice as long as cube A. Which of the following is true?

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(C). Density depends only on the material, so it is the same. Volume of a cube is side³. Doubling the side gives 2³ = 8 times the volume. Same density means mass = ρV is also 8 times.

(A) and (D) wrongly let density change with size. (B) forgets that volume grows with the cube of the side.

3. A hiker of mass 60 kg stands on snow. In boots, the total area touching the snow is 0.040 m². In snowshoes, it is 0.20 m². How does the pressure on the snow in boots compare to the pressure in snowshoes?

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(C). The force is the hiker's weight, mg = 60 × 9.8 = 588 N, in both cases. P = F⊥ / A. Boots: 588 / 0.040 = 14 700 Pa. Snowshoes: 588 / 0.20 = 2940 Pa. Ratio = 14 700 / 2940 = 5.

(A) flips the ratio. (B) thinks the same force means the same pressure. (D) squares the area ratio, but pressure depends on area to the first power.

4. A diver is 15 m below the surface of a freshwater lake. The air above the lake is at 1.0 × 10⁵ Pa. What is the absolute pressure on the diver, to two significant figures?

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(C). Gauge pressure: ρgh = 1000 × 9.8 × 15 = 147 000 Pa = 1.47 × 10⁵ Pa. Absolute pressure: P = P₀ + ρgh = 1.0 × 10⁵ + 1.47 × 10⁵ = 2.47 × 10⁵ Pa ≈ 2.5 × 10⁵ Pa.

(A) is just the air pressure. (B) is the gauge pressure only (it forgets the air on top). (D) adds the air pressure twice.

5. Three open containers sit on a table. Container X is a wide bowl. Container Y is a tall, narrow vase. Container Z is a cone that gets wider toward the top. Each holds water to the same depth, 0.30 m. Rank the water pressure at the bottom of each container.

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(D). Pressure in a still fluid depends only on depth: P = P₀ + ρgh. Same fluid, same depth, same air above, so the same pressure: Pgauge = 1000 × 9.8 × 0.30 = 2940 Pa in all three.

The other choices think the amount of water or the shape matters. The total force on each bottom can be different (F = PA, and the bottoms have different areas), but the pressure is the same.

6. A pressure sensor is lowered slowly into an open tank of water. Which description best matches a graph of the absolute pressure P (vertical axis) against depth h (horizontal axis)?

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(B). P = P₀ + ρgh has the form y = b + mx. The intercept is P₀ (air pressure at the surface) and the slope is ρg, which is constant because an ideal fluid is incompressible.

(A) is the graph of gauge pressure, not absolute. (C) would need the density to grow with depth. (D) mixes up "density is constant" with "pressure is constant".

7. A 2.0 kg metal block with a volume of 5.0 × 10⁻⁴ m³ hangs from a spring scale. The block is fully under water and does not touch the bottom. What does the scale read?

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(B). Forces on the block: tension T up, buoyant force Fb up, weight mg down. At rest: T + Fb = mg.

Weight: mg = 2.0 × 9.8 = 19.6 N. Buoyant force: Fb = ρVg = 1000 × 5.0 × 10⁻⁴ × 9.8 = 4.9 N. So T = 19.6 − 4.9 = 14.7 N.

(A) is the buoyant force only. (C) is the weight in air. (D) adds the buoyant force instead of subtracting.

8. A wooden block floats in fresh water. It is moved to a tank of oil (density 800 kg/m³), where it still floats. Compared with floating in water, what happens to the buoyant force on the block and to the volume of the block below the surface?

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(B). A floating object is in equilibrium, so Fb = mg in both liquids. The weight did not change, so the buoyant force did not change. Since Fb = ρfluidVsubg and ρfluid is smaller in oil, Vsub must be larger. The block sits lower.

(A) and (D) forget that floating means Fb equals the weight. (C) gets the direction wrong: a less dense liquid has to be pushed aside more.

9. A beaker of water sits on a scale that reads 10.0 N. A rock that weighs 3.0 N and has a volume of 1.0 × 10⁻⁴ m³ hangs from a string held by a student. The rock is lowered until it is fully under water, but it does not touch the beaker. What does the scale read now?

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(B). The water pushes up on the rock with Fb = ρVg = 1000 × 1.0 × 10⁻⁴ × 9.8 = 0.98 N. By Newton's third law, the rock pushes down on the water with 0.98 N. That push reaches the scale. New reading: 10.0 + 0.98 = 10.98 N ≈ 11.0 N.

(A) ignores the third-law pair. (D) adds the whole weight of the rock, which would be true only if the rock rested on the bottom (or the string were cut). (C) is 13.0 − 0.98, mixing the two ideas.

10. A cork of mass 0.010 kg and volume 4.0 × 10⁻⁵ m³ is held fully under water by a thin string tied to the bottom of a tank. What is the tension in the string?

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(B). Forces on the cork: Fb up, weight down, tension down (the string pulls it toward the bottom). At rest: Fb = mg + T.

Fb = 1000 × 4.0 × 10⁻⁵ × 9.8 = 0.392 N. mg = 0.010 × 9.8 = 0.098 N. T = 0.392 − 0.098 = 0.294 N ≈ 0.29 N.

(A) is the weight. (C) is the buoyant force. (D) adds them, as if tension pointed up.

11. A solid aluminum sphere and a solid lead sphere have the same volume. Both are fully under water, held by strings. How do the buoyant forces on them compare?

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(C). Fb = ρfluidVsubg. Same water, same submerged volume, so the same buoyant force. The buoyant force depends on the fluid displaced, not on what the object is made of.

(A), (B) and (D) all wrongly link buoyant force to the object's own mass or density. The masses only change the string tensions.

12. Water flows through a garden hose with a cross-sectional area of 3.0 cm² at 2.0 m/s. It then leaves through a nozzle with an area of 0.50 cm². How fast does the water leave the nozzle?

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(C). Continuity: A₁v₁ = A₂v₂. v₂ = A₁v₁ / A₂ = (3.0 cm² × 2.0 m/s) / 0.50 cm² = 12 m/s. The area units cancel, so there is no need to convert cm² to m².

(A) flips the ratio (narrow should mean faster). (B) thinks speed stays the same. (D) squares the area ratio; that is what you do with a radius ratio, not an area ratio.

13. Water flows at 1.5 m/s through a pipe with an inside radius of 1.0 cm. About how long does it take to fill a 0.010 m³ (10 liter) bucket?

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(C). Area: A = πr² = π(0.010 m)² = 3.14 × 10⁻⁴ m². Volume flow rate: V/t = Av = 3.14 × 10⁻⁴ × 1.5 = 4.71 × 10⁻⁴ m³/s. Time: t = 0.010 / 4.71 × 10⁻⁴ = 21 s.

(A) and (B) are powers-of-ten slips from converting cm to m. (D) uses the diameter (2.0 cm) where the radius belongs, which makes the area 4 times too big.

14. Water flows through a horizontal pipe. At point 1 the speed is 2.0 m/s and the absolute pressure is 1.80 × 10⁵ Pa. At point 2 the pipe is narrower and the speed is 6.0 m/s. What is the pressure at point 2?

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(B). Bernoulli with y₁ = y₂: P₁ + ½ρv₁² = P₂ + ½ρv₂². So P₂ = P₁ − ½ρ(v₂² − v₁²) = 1.80 × 10⁵ − ½(1000)(36 − 4) = 1.80 × 10⁵ − 16 000 = 1.64 × 10⁵ Pa.

(A) adds the change (faster flow means lower pressure). (C) uses only v₂² and forgets v₁. (D) forgets the ½.

15. A large open tank has a small hole in its side 1.25 m below the water surface. About how fast does water leave the hole?

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(B). Torricelli: v = √(2gh) = √(2 × 9.8 × 1.25) = √24.5 = 4.9 m/s. It is the same speed a ball gets falling 1.25 m. Both the surface and the hole are open to air, and the surface barely moves because the tank is large.

(A) forgets the 2: √(9.8 × 1.25). (C) is gh with no square root. (D) is 2gh with no square root.

16. Water flows from left to right through a horizontal pipe with three sections. Section 1 has area A, section 2 has area 2A, and section 3 has area A/2. Rank the pressures in the three sections from highest to lowest.

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(B). Continuity: v = (volume flow rate)/A, so the speed is lowest in the widest part: v₂ < v₁ < v₃. Bernoulli at the same height: where the speed is lower, the pressure is higher. So P₂ > P₁ > P₃.

(A) is the ranking of speeds, not pressures. (C) ignores Bernoulli. (D) assumes pressure must drop in the direction of flow no matter what; in an ideal fluid, it rises again where the pipe widens.

17. A solid cube denser than water hangs from a string and is lowered slowly into a deep tank, bottom face first, at constant speed. Which description best matches a graph of the buoyant force on the cube against the depth of the cube's bottom face?

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(A). Fb = ρgVsub. While the cube goes in, Vsub = (face area) × (depth), which grows linearly. Once it is fully under, Vsub stops growing, so Fb is constant.

(B) is the classic trap: pressure on the top and on the bottom both grow with depth, but their difference stays the same. (C) and (D) do not match Vsub.

18. In AP Physics 1, fluids are assumed to be "ideal" unless a question says otherwise. Which statement describes an ideal fluid?

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(A). An ideal fluid is incompressible (its density stays the same under pressure) and has no viscosity. That is why ρ is a constant in P = P₀ + ρgh and why Bernoulli's equation conserves energy.

(B) gets compression wrong. (C) adds a false rule: ideal fluids can have any density. (D) is false: Bernoulli is all about ideal fluids that flow.

Section II: Free response (5 questions)

Show your work: write the equation you start from, then the numbers with units. On the real exam, a correct answer with no reasoning earns little. Points are shown so you can grade yourself.

1. Mathematical Routines: the cargo raft 10 points

A raft is a solid block of wood with density ρr = 650 kg/m³. It is 2.0 m long, 1.5 m wide and 0.20 m thick. It floats flat in a freshwater lake (ρw = 1000 kg/m³).

  1. Derive an expression for the depth d of the raft below the waterline when it carries no cargo. Write it in terms of ρr, ρw, the thickness t, and physical constants as needed. Then find the number.
  2. Calculate the largest mass of cargo the raft can carry before water reaches its top surface.
  3. The raft, with no cargo, is towed into seawater (density 1025 kg/m³). Does it float higher, lower, or at the same level as in fresh water? Justify your answer.
  4. With the largest cargo from part (b), the bottom of the raft is 0.20 m below the lake surface. Calculate the gauge pressure on the bottom of the raft two different ways, and show they agree.
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(a) 3 points. Floating means equilibrium, so Fb = mrg. (1 pt) With footprint area A: ρw(A d)g = ρr(A t)g. (1 pt) A and g cancel: d = (ρr/ρw) t = (650/1000)(0.20 m) = 0.13 m. (1 pt)

(b) 3 points. Raft volume V = 2.0 × 1.5 × 0.20 = 0.60 m³. Raft mass = 650 × 0.60 = 390 kg. (1 pt) When the raft is just fully under, Fb = ρwVg, which holds up a total mass of ρwV = 1000 × 0.60 = 600 kg. (1 pt) Largest cargo = 600 − 390 = 210 kg. (1 pt)

(c) 2 points. Higher. (1 pt) The buoyant force must still equal the same weight. Seawater is denser, so a smaller submerged volume gives that force: d = (650/1025)(0.20) = 0.127 m, less than 0.13 m. (1 pt)

(d) 2 points. From depth: Pgauge = ρgh = 1000 × 9.8 × 0.20 = 1960 Pa. (1 pt) From force: total weight = 600 kg × 9.8 = 5880 N spread over A = 2.0 × 1.5 = 3.0 m², so P = F/A = 5880 / 3.0 = 1960 Pa. They agree because the water's upward push on the bottom is exactly what holds up the raft and cargo. (1 pt)

2. Translation Between Representations: lowering a metal block 12 points

A metal block of mass m = 0.80 kg has a square bottom face of area A = 2.0 × 10⁻³ m² and height H = 0.050 m. It hangs from a spring scale. A student lowers it slowly into a beaker of water that sits on a balance. Let y be the depth of the block's bottom face below the water surface. The block never touches the beaker.

  1. The block is half under water (y = H/2). List every force on the block, give its direction, and say which object exerts it. State how the sizes compare (which is largest, and how the others add up to it).
  2. Choose the description that best matches the spring scale reading T against y, from y = 0 to y = 2H:
    (i) constant; (ii) falls in a straight line from y = 0 to y = H, then constant; (iii) falls in a straight line the whole way; (iv) rises in a straight line, then constant.
    Then choose, from the same list, the shape of the balance reading against y.
  3. Write an equation for T in terms of y for 0 ≤ y ≤ H, using m, ρw, A, g and y. Show that the slope of your equation matches your graph choice in (b).
  4. Calculate the scale reading when the block is fully under water, and the slope of the T vs y graph (with units).
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(a) 3 points. Weight mg, down, exerted by Earth. Tension T, up, exerted by the string (scale). Buoyant force Fb, up, exerted by the water. (2 pts: all three with correct direction and source) The weight is largest, and T + Fb = mg because the block is at rest (moving slowly at constant speed). (1 pt)

(b) 4 points. Scale: (ii). (1 pt) While the block goes in, Vsub = Ay grows linearly, so Fb grows linearly and T = mg − Fb falls linearly. After y = H, Vsub stops changing, so T is constant. (1 pt) Balance: (iv). (1 pt) By Newton's third law, the block pushes down on the water with a force equal to Fb, so the balance reading rises by the same amount T falls. (1 pt)

(c) 3 points. T = mg − Fb = mg − ρw(Ay)g. (1 pt) This is a line T = mg − (ρwAg) y with intercept mg and slope −ρwAg. (1 pt) The slope is negative and constant, which matches a straight line going down in (ii). (1 pt)

(d) 2 points. V = AH = 2.0 × 10⁻³ × 0.050 = 1.0 × 10⁻⁴ m³. Fb = 1000 × 1.0 × 10⁻⁴ × 9.8 = 0.98 N. mg = 0.80 × 9.8 = 7.84 N. T = 7.84 − 0.98 = 6.86 N. (1 pt) Slope = −ρwAg = −1000 × 2.0 × 10⁻³ × 9.8 = −19.6 N/m. Check: 19.6 N/m × 0.050 m = 0.98 N, the full drop. (1 pt)

3. Experimental Design and Analysis: density of an unknown liquid 10 points

A student wants to find the density of an unknown liquid. She has a large beaker of the liquid, a spring scale, string, a ruler, and a set of five metal cylinders of different known volumes. All of the cylinders sink.

  1. Describe a procedure she could use to collect data that lets her find the density of the liquid with a graph. Say what she measures, and with which tool.
  2. Say what quantities she should plot on each axis to get a straight line, and how the density is found from that line.
  3. Here is her data. ΔF is the weight in air minus the scale reading when the cylinder is fully in the liquid.
    V (× 10⁻⁵ m³)ΔF (N)
    2.00.16
    4.00.30
    6.00.48
    8.00.62
    10.00.79
    Use the data to find the slope of a best-fit line, and from it the density of the liquid.
  4. In one trial, a cylinder rested on the bottom of the beaker while the scale was read. Would that make ΔF for that trial too large or too small? Explain.
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(a) 3 points. For each cylinder: hang it from the spring scale in air and record the weight W. (1 pt) Lower it until it is completely under the liquid, not touching the sides or bottom, and record the scale reading T. (1 pt) Compute ΔF = W − T, which equals the buoyant force. Repeat for all five cylinders (each with its known volume V). (1 pt)

(b) 2 points. Plot ΔF (vertical) against V (horizontal). (1 pt) Because ΔF = Fb = ρliquidgV, the line passes through the origin with slope ρliquidg, so ρliquid = slope / g. (1 pt)

(c) 3 points. A best-fit line through the points (a least-squares fit gives 0.079 N per 1.0 × 10⁻⁵ m³) has slope ≈ 7.9 × 10³ N/m³. A line drawn by eye from (0, 0) to (10.0 × 10⁻⁵ m³, 0.79 N) gives the same: 0.79 / 1.0 × 10⁻⁴ = 7.9 × 10³ N/m³. (2 pts: slope from the line, not from one data point, with units) ρ = 7.9 × 10³ / 9.8 ≈ 810 kg/m³ (any answer from 750 to 850 kg/m³ with correct method earns the point). (1 pt) That is close to the density of a light oil.

(d) 2 points. Too large. (1 pt) The bottom of the beaker pushes up on the cylinder too (a normal force), so the scale reads less than it should. A smaller T makes ΔF = W − T bigger, so that point sits above the line. (1 pt)

4. Qualitative/Quantitative Translation: where does the jet land? 8 points

A large open tank stands on level ground. The water surface is at a height D above the ground. A small hole can be drilled in the side at a depth h below the water surface. The tank is so wide that the surface barely moves. A student claims: "The lower the hole, the faster the water comes out, so the lowest hole always sprays the farthest."

  1. Without equations, explain in a short paragraph why the student's claim is not correct. Mention both the speed of the water and the time it spends in the air.
  2. Derive an expression for the horizontal distance x from the tank wall to where the jet lands, in terms of h, D and physical constants.
  3. Explain how your expression from (b) supports your paragraph in (a). At what depth h does the jet land farthest?
  4. D = 1.8 m. Calculate x for a hole at h = 0.50 m and for a hole at h = 0.90 m.
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(a) 2 points. A deeper hole has more water above it, so the pressure there is higher and the water leaves faster. (1 pt) But a deeper hole is also closer to the ground, so the water falls a shorter height and spends less time in the air before landing. Distance depends on both speed and time, so a very low hole (fast but almost no time in the air) can land close. The claim ignores the time. (1 pt)

(b) 3 points. Exit speed from Torricelli: v = √(2gh). (1 pt) The hole is a height (D − h) above the ground. The water leaves horizontally, so the fall time comes from D − h = ½gt², t = √(2(D − h)/g). (1 pt) Horizontal distance: x = vt = √(2gh) × √(2(D − h)/g) = 2√(h(D − h)). (1 pt)

(c) 2 points. In x = 2√(h(D − h)), the factor h grows as the hole gets deeper (faster jet), but the factor (D − h) shrinks (less fall time). When h → D, x → 0, so the lowest hole does not spray farthest. (1 pt) The product h(D − h) is largest when h = D/2, the middle of the water column. (1 pt)

(d) 1 point. h = 0.50 m: x = 2√(0.50 × 1.3) = 2√0.65 = 1.6 m. h = 0.90 m (the middle): x = 2√(0.90 × 0.90) = 1.8 m, the largest possible. (A hole at h = 1.3 m also gives 1.6 m: the expression is symmetric about D/2.) (1 pt) Note that g cancels: the landing spot is the same on any planet.

5. Mathematical Routines (extra): water up to the second floor 6 points

Water enters a house through a basement pipe of inside radius 1.0 cm at a speed of 1.5 m/s and an absolute pressure of 3.0 × 10⁵ Pa. The pipe rises 5.0 m to a second-floor pipe of inside radius 0.50 cm. Treat the water as an ideal fluid.

  1. Calculate the speed of the water in the second-floor pipe.
  2. Calculate the absolute pressure in the second-floor pipe.
  3. Calculate the volume flow rate in the second-floor pipe.
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(a) 2 points. Continuity: A₁v₁ = A₂v₂, and A = πr², so v₂ = v₁(r₁/r₂)² = 1.5 × (1.0/0.50)² = 1.5 × 4 = 6.0 m/s. (1 pt for continuity, 1 pt for squaring the radius ratio)

(b) 3 points. Bernoulli: P₁ + ρgy₁ + ½ρv₁² = P₂ + ρgy₂ + ½ρv₂², with y₁ = 0, y₂ = 5.0 m. (1 pt) P₂ = P₁ − ρg(y₂ − y₁) − ½ρ(v₂² − v₁²) = 3.0 × 10⁵ − (1000)(9.8)(5.0) − ½(1000)(36 − 2.25) (1 pt) = 300 000 − 49 000 − 16 875 = 234 125 Pa ≈ 2.3 × 10⁵ Pa. (1 pt) Both going up and speeding up cost pressure.

(c) 1 point. V/t = A₂v₂ = π(0.0050)² × 6.0 = 7.85 × 10⁻⁵ × 6.0 = 4.7 × 10⁻⁴ m³/s. Check: the basement gives π(0.010)² × 1.5 = 4.7 × 10⁻⁴ m³/s, the same, as continuity demands.

Scoring guide

Weak spot? Go back to the topic: 8.1 Density, 8.2 Pressure, 8.3 Fluids and Newton's laws, 8.4 Fluids and conservation laws, or play the Unit 8 games.